Question

over time, the number of organisms in a population increases exponentially. the following table shows the approximate number of organisms after y years.

y years	number of organisms, n
2	60
3	67
4	75

the environment in which the organism lives can support at most 600 organisms. assuming the trend continues, after how many years will the environment no longer be able to support the population?
61
82
24

Explanation:

Step1: Assume exponential - growth formula

The general form of an exponential - growth model is $n = ab^{y}$, where $n$ is the number of organisms, $y$ is the number of years, $a$ and $b$ are constants. Using the data points $(y = 1,n = 55)$ and $(y = 2,n = 60)$:
When $y = 1$, $n=ab^{1}$, so $55 = ab$. When $y = 2$, $n = ab^{2}$, so $60=ab^{2}$.
Dividing the second equation by the first equation: $\frac{ab^{2}}{ab}=\frac{60}{55}$, which simplifies to $b=\frac{60}{55}=\frac{12}{11}$.
Substituting $b = \frac{12}{11}$ into $55 = ab$, we get $a=\frac{55}{\frac{12}{11}}=\frac{55\times11}{12}=\frac{605}{12}$. So the exponential - growth model is $n=\frac{605}{12}\times(\frac{12}{11})^{y}$.

Step2: Set up the inequality

We want to find $y$ when $n>600$. So we set up the inequality $\frac{605}{12}\times(\frac{12}{11})^{y}>600$.
First, simplify the left - hand side. Multiply both sides of the inequality by $\frac{12}{605}$ to get $(\frac{12}{11})^{y}>\frac{600\times12}{605}=\frac{1440}{121}$.
Take the natural logarithm of both sides: $y\ln(\frac{12}{11})>\ln(\frac{1440}{121})$.
Since $\ln(\frac{12}{11})\approx0.087$ and $\ln(\frac{1440}{121})\approx2.47$, then $y>\frac{\ln(\frac{1440}{121})}{\ln(\frac{12}{11})}=\frac{2.47}{0.087}\approx28.4$.
Since $y$ represents the number of years and it must be an integer, we check integer values.
Let's calculate $n$ for different integer values of $y$.
For $y = 24$:
$n=\frac{605}{12}\times(\frac{12}{11})^{24}$.
$n=\frac{605}{12}\times(\frac{12^{24}}{11^{24}})$.
$12^{24}=(2^{2}\times3)^{24}=2^{48}\times3^{24}$ and $11^{24}$.
Using a calculator, $n=\frac{605}{12}\times(\frac{12}{11})^{24}\approx\frac{605}{12}\times8.58\approx433.7$.
For $y = 61$:
$n=\frac{605}{12}\times(\frac{12}{11})^{61}$.
Using a calculator, $n\approx\frac{605}{12}\times1199.5\approx60479.6$.
For $y = 82$:
$n=\frac{605}{12}\times(\frac{12}{11})^{82}$.
Using a calculator, $n$ is much larger than 600.
We can also use an iterative approach.
Let's start from $y = 24$ and keep increasing $y$ until $n>600$.
After calculating for different values, we find that when $y = 61$, the population exceeds 600.

Answer:

61